$ time foo
real 0m0.003s
user 0m0.000s
sys 0m0.004s
$
What do 'real', 'user' and 'sys' mean in the output of time?
Which one is meaningful when benchmarking my app?
Real, User and Sys process time statistics
One of these things is not like the other. Real refers to actual elapsed time; User and Sys refer to CPU time used only by the process.
Real is wall clock time - time from start to finish of the call. This is all elapsed time including time slices used by other processes and time the process spends blocked (for example if it is waiting for I/O to complete).
User is the amount of CPU time spent in user-mode code (outside the kernel) within the process. This is only actual CPU time used in executing the process. Other processes and time the process spends blocked do not count towards this figure.
Sys is the amount of CPU time spent in the kernel within the process. This means executing CPU time spent in system calls within the kernel, as opposed to library code, which is still running in user-space. Like 'user', this is only CPU time used by the process. See below for a brief description of kernel mode (also known as 'supervisor' mode) and the system call mechanism.
User+Sys
will tell you how much actual CPU time your process used. Note that this is across all CPUs, so if the process has multiple threads (and this process is running on a computer with more than one processor) it could potentially exceed the wall clock time reported by Real
(which usually occurs). Note that in the output these figures include the User
and Sys
time of all child processes (and their descendants) as well when they could have been collected, e.g. by wait(2)
or waitpid(2)
, although the underlying system calls return the statistics for the process and its children separately.
Origins of the statistics reported by time (1)
The statistics reported by time
are gathered from various system calls. 'User' and 'Sys' come from wait (2)
(POSIX) or times (2)
(POSIX), depending on the particular system. 'Real' is calculated from a start and end time gathered from the gettimeofday (2)
call. Depending on the version of the system, various other statistics such as the number of context switches may also be gathered by time
.
On a multi-processor machine, a multi-threaded process or a process forking children could have an elapsed time smaller than the total CPU time - as different threads or processes may run in parallel. Also, the time statistics reported come from different origins, so times recorded for very short running tasks may be subject to rounding errors, as the example given by the original poster shows.
A brief primer on Kernel vs. User mode
On Unix, or any protected-memory operating system, 'Kernel' or 'Supervisor' mode refers to a privileged mode that the CPU can operate in. Certain privileged actions that could affect security or stability can only be done when the CPU is operating in this mode; these actions are not available to application code. An example of such an action might be manipulation of the MMU to gain access to the address space of another process. Normally, user-mode code cannot do this (with good reason), although it can request shared memory from the kernel, which could be read or written by more than one process. In this case, the shared memory is explicitly requested from the kernel through a secure mechanism and both processes have to explicitly attach to it in order to use it.
The privileged mode is usually referred to as 'kernel' mode because the kernel is executed by the CPU running in this mode. In order to switch to kernel mode you have to issue a specific instruction (often called a trap) that switches the CPU to running in kernel mode and runs code from a specific location held in a jump table. For security reasons, you cannot switch to kernel mode and execute arbitrary code - the traps are managed through a table of addresses that cannot be written to unless the CPU is running in supervisor mode. You trap with an explicit trap number and the address is looked up in the jump table; the kernel has a finite number of controlled entry points.
The 'system' calls in the C library (particularly those described in Section 2 of the man pages) have a user-mode component, which is what you actually call from your C program. Behind the scenes, they may issue one or more system calls to the kernel to do specific services such as I/O, but they still also have code running in user-mode. It is also quite possible to directly issue a trap to kernel mode from any user space code if desired, although you may need to write a snippet of assembly language to set up the registers correctly for the call.
More about 'sys'
There are things that your code cannot do from user mode - things like allocating memory or accessing hardware (HDD, network, etc.). These are under the supervision of the kernel, and it alone can do them. Some operations like malloc
orfread
/fwrite
will invoke these kernel functions and that then will count as 'sys' time. Unfortunately it's not as simple as "every call to malloc will be counted in 'sys' time". The call to malloc
will do some processing of its own (still counted in 'user' time) and then somewhere along the way it may call the function in kernel (counted in 'sys' time). After returning from the kernel call, there will be some more time in 'user' and then malloc
will return to your code. As for when the switch happens, and how much of it is spent in kernel mode... you cannot say. It depends on the implementation of the library. Also, other seemingly innocent functions might also use malloc
and the like in the background, which will again have some time in 'sys' then.
To expand on the accepted answer, I just wanted to provide another reason why real
≠ user
+ sys
.
Keep in mind that real
represents actual elapsed time, while user
and sys
values represent CPU execution time. As a result, on a multicore system, the user
and/or sys
time (as well as their sum) can actually exceed the real time. For example, on a Java app I'm running for class I get this set of values:
real 1m47.363s
user 2m41.318s
sys 0m4.013s
• real: The actual time spent in running the process from start to finish, as if it was measured by a human with a stopwatch
• user: The cumulative time spent by all the CPUs during the computation
• sys: The cumulative time spent by all the CPUs during system-related tasks such as memory allocation.
Notice that sometimes user + sys might be greater than real, as multiple processors may work in parallel.
Minimal runnable POSIX C examples
To make things more concrete, I want to exemplify a few extreme cases of time
with some minimal C test programs.
All programs can be compiled and run with:
gcc -ggdb3 -o main.out -pthread -std=c99 -pedantic-errors -Wall -Wextra main.c
time ./main.out
and have been tested in Ubuntu 18.10, GCC 8.2.0, glibc 2.28, Linux kernel 4.18, ThinkPad P51 laptop, Intel Core i7-7820HQ CPU (4 cores / 8 threads), 2x Samsung M471A2K43BB1-CRC RAM (2x 16GiB).
sleep
syscall
Non-busy sleep as done by the sleep
syscall only counts in real
, but not for user
or sys
.
For example, a program that sleeps for a second:
#define _XOPEN_SOURCE 700
#include <stdlib.h>
#include <unistd.h>
int main(void) {
sleep(1);
return EXIT_SUCCESS;
}
outputs something like:
real 0m1.003s
user 0m0.001s
sys 0m0.003s
The same holds for programs blocked on IO becoming available.
For example, the following program waits for the user to enter a character and press enter:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
printf("%c\n", getchar());
return EXIT_SUCCESS;
}
And if you wait for about one second, it outputs just like the sleep example something like:
real 0m1.003s
user 0m0.001s
sys 0m0.003s
For this reason time
can help you distinguish between CPU and IO bound programs: What do the terms "CPU bound" and "I/O bound" mean?
Multiple threads
The following example does niters
iterations of useless purely CPU-bound work on nthreads
threads:
#define _XOPEN_SOURCE 700
#include <assert.h>
#include <inttypes.h>
#include <pthread.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
uint64_t niters;
void* my_thread(void *arg) {
uint64_t *argument, i, result;
argument = (uint64_t *)arg;
result = *argument;
for (i = 0; i < niters; ++i) {
result = (result * result) - (3 * result) + 1;
}
*argument = result;
return NULL;
}
int main(int argc, char **argv) {
size_t nthreads;
pthread_t *threads;
uint64_t rc, i, *thread_args;
/* CLI args. */
if (argc > 1) {
niters = strtoll(argv[1], NULL, 0);
} else {
niters = 1000000000;
}
if (argc > 2) {
nthreads = strtoll(argv[2], NULL, 0);
} else {
nthreads = 1;
}
threads = malloc(nthreads * sizeof(*threads));
thread_args = malloc(nthreads * sizeof(*thread_args));
/* Create all threads */
for (i = 0; i < nthreads; ++i) {
thread_args[i] = i;
rc = pthread_create(
&threads[i],
NULL,
my_thread,
(void*)&thread_args[i]
);
assert(rc == 0);
}
/* Wait for all threads to complete */
for (i = 0; i < nthreads; ++i) {
rc = pthread_join(threads[i], NULL);
assert(rc == 0);
printf("%" PRIu64 " %" PRIu64 "\n", i, thread_args[i]);
}
free(threads);
free(thread_args);
return EXIT_SUCCESS;
}
Then we plot wall, user and sys as a function of the number of threads for a fixed 10^10 iterations on my 8 hyperthread CPU:
From the graph, we see that:
for a CPU intensive single core application, wall and user are about the same
for 2 cores, user is about 2x wall, which means that the user time is counted across all threads.
user basically doubled, and while wall stayed the same.
this continues up to 8 threads, which matches my number of hyperthreads in my computer.
After 8, wall starts to increase as well, because we don't have any extra CPUs to put more work in a given amount of time!
The ratio plateaus at this point.
Note that this graph is only so clear and simple because the work is purely CPU-bound: if it were memory bound, then we would get a fall in performance much earlier with less cores because the memory accesses would be a bottleneck as shown at What do the terms "CPU bound" and "I/O bound" mean?
Quickly checking that wall < user is a simple way to determine that a program is multithreaded, and the closer that ratio is to the number of cores, the more effective the parallelization is, e.g.:
Sys heavy work with sendfile
The heaviest sys workload I could come up with was to use the sendfile
, which does a file copy operation on kernel space: Copy a file in a sane, safe and efficient way
So I imagined that this in-kernel memcpy
will be a CPU intensive operation.
First I initialize a large 10GiB random file with:
dd if=/dev/urandom of=sendfile.in.tmp bs=1K count=10M
Then run the code:
#define _GNU_SOURCE
#include <assert.h>
#include <fcntl.h>
#include <stdlib.h>
#include <sys/sendfile.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
int main(int argc, char **argv) {
char *source_path, *dest_path;
int source, dest;
struct stat stat_source;
if (argc > 1) {
source_path = argv[1];
} else {
source_path = "sendfile.in.tmp";
}
if (argc > 2) {
dest_path = argv[2];
} else {
dest_path = "sendfile.out.tmp";
}
source = open(source_path, O_RDONLY);
assert(source != -1);
dest = open(dest_path, O_WRONLY | O_CREAT | O_TRUNC, S_IRUSR | S_IWUSR);
assert(dest != -1);
assert(fstat(source, &stat_source) != -1);
assert(sendfile(dest, source, 0, stat_source.st_size) != -1);
assert(close(source) != -1);
assert(close(dest) != -1);
return EXIT_SUCCESS;
}
which gives basically mostly system time as expected:
real 0m2.175s
user 0m0.001s
sys 0m1.476s
I was also curious to see if time
would distinguish between syscalls of different processes, so I tried:
time ./sendfile.out sendfile.in1.tmp sendfile.out1.tmp &
time ./sendfile.out sendfile.in2.tmp sendfile.out2.tmp &
And the result was:
real 0m3.651s
user 0m0.000s
sys 0m1.516s
real 0m4.948s
user 0m0.000s
sys 0m1.562s
The sys time is about the same for both as for a single process, but the wall time is larger because the processes are competing for disk read access likely.
So it seems that it does in fact account for which process started a given kernel work.
Bash source code
When you do just time <cmd>
on Ubuntu, it use the Bash keyword as can be seen from:
type time
which outputs:
time is a shell keyword
So we grep source in the Bash 4.19 source code for the output string:
git grep '"user\b'
which leads us to execute_cmd.c function time_command
, which uses:
gettimeofday()
and getrusage()
if both are availabletimes()
otherwiseall of which are Linux system calls and POSIX functions.
GNU Coreutils source code
If we call it as:
/usr/bin/time
then it uses the GNU Coreutils implementation.
This one is a bit more complex, but the relevant source seems to be at resuse.c and it does:
a non-POSIX BSD wait3
call if that is available
times
and gettimeofday
otherwise
1: https://i.stack.imgur.com/qAfEe.png**Minimal runnable POSIX C examples**
To make things more concrete, I want to exemplify a few extreme cases of time
with some minimal C test programs.
All programs can be compiled and run with:
gcc -ggdb3 -o main.out -pthread -std=c99 -pedantic-errors -Wall -Wextra main.c
time ./main.out
and have been tested in Ubuntu 18.10, GCC 8.2.0, glibc 2.28, Linux kernel 4.18, ThinkPad P51 laptop, Intel Core i7-7820HQ CPU (4 cores / 8 threads), 2x Samsung M471A2K43BB1-CRC RAM (2x 16GiB).
sleep
Non-busy sleep does not count in either user
or sys
, only real
.
For example, a program that sleeps for a second:
#define _XOPEN_SOURCE 700
#include <stdlib.h>
#include <unistd.h>
int main(void) {
sleep(1);
return EXIT_SUCCESS;
}
outputs something like:
real 0m1.003s
user 0m0.001s
sys 0m0.003s
The same holds for programs blocked on IO becoming available.
For example, the following program waits for the user to enter a character and press enter:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
printf("%c\n", getchar());
return EXIT_SUCCESS;
}
And if you wait for about one second, it outputs just like the sleep example something like:
real 0m1.003s
user 0m0.001s
sys 0m0.003s
For this reason time
can help you distinguish between CPU and IO bound programs: What do the terms "CPU bound" and "I/O bound" mean?
Multiple threads
The following example does niters
iterations of useless purely CPU-bound work on nthreads
threads:
#define _XOPEN_SOURCE 700
#include <assert.h>
#include <inttypes.h>
#include <pthread.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
uint64_t niters;
void* my_thread(void *arg) {
uint64_t *argument, i, result;
argument = (uint64_t *)arg;
result = *argument;
for (i = 0; i < niters; ++i) {
result = (result * result) - (3 * result) + 1;
}
*argument = result;
return NULL;
}
int main(int argc, char **argv) {
size_t nthreads;
pthread_t *threads;
uint64_t rc, i, *thread_args;
/* CLI args. */
if (argc > 1) {
niters = strtoll(argv[1], NULL, 0);
} else {
niters = 1000000000;
}
if (argc > 2) {
nthreads = strtoll(argv[2], NULL, 0);
} else {
nthreads = 1;
}
threads = malloc(nthreads * sizeof(*threads));
thread_args = malloc(nthreads * sizeof(*thread_args));
/* Create all threads */
for (i = 0; i < nthreads; ++i) {
thread_args[i] = i;
rc = pthread_create(
&threads[i],
NULL,
my_thread,
(void*)&thread_args[i]
);
assert(rc == 0);
}
/* Wait for all threads to complete */
for (i = 0; i < nthreads; ++i) {
rc = pthread_join(threads[i], NULL);
assert(rc == 0);
printf("%" PRIu64 " %" PRIu64 "\n", i, thread_args[i]);
}
free(threads);
free(thread_args);
return EXIT_SUCCESS;
}
Then we plot wall, user and sys as a function of the number of threads for a fixed 10^10 iterations on my 8 hyperthread CPU:
From the graph, we see that:
for a CPU intensive single core application, wall and user are about the same
for 2 cores, user is about 2x wall, which means that the user time is counted across all threads.
user basically doubled, and while wall stayed the same.
this continues up to 8 threads, which matches my number of hyperthreads in my computer.
After 8, wall starts to increase as well, because we don't have any extra CPUs to put more work in a given amount of time!
The ratio plateaus at this point.
Note that this graph is only so clear and simple because the work is purely CPU-bound: if it were memory bound, then we would get a fall in performance much earlier with less cores because the memory accesses would be a bottleneck as shown at What do the terms "CPU bound" and "I/O bound" mean?
Quickly checking that wall < user is a simple way to determine that a program is multithreaded, and the closer that ratio is to the number of cores, the more effective the parallelization is, e.g.:
Sys heavy work with sendfile
The heaviest sys workload I could come up with was to use the sendfile
, which does a file copy operation on kernel space: Copy a file in a sane, safe and efficient way
So I imagined that this in-kernel memcpy
will be a CPU intensive operation.
First I initialize a large 10GiB random file with:
dd if=/dev/urandom of=sendfile.in.tmp bs=1K count=10M
Then run the code:
#define _GNU_SOURCE
#include <assert.h>
#include <fcntl.h>
#include <stdlib.h>
#include <sys/sendfile.h>
#include <sys/stat.h>
#include <sys/types.h>
#include <unistd.h>
int main(int argc, char **argv) {
char *source_path, *dest_path;
int source, dest;
struct stat stat_source;
if (argc > 1) {
source_path = argv[1];
} else {
source_path = "sendfile.in.tmp";
}
if (argc > 2) {
dest_path = argv[2];
} else {
dest_path = "sendfile.out.tmp";
}
source = open(source_path, O_RDONLY);
assert(source != -1);
dest = open(dest_path, O_WRONLY | O_CREAT | O_TRUNC, S_IRUSR | S_IWUSR);
assert(dest != -1);
assert(fstat(source, &stat_source) != -1);
assert(sendfile(dest, source, 0, stat_source.st_size) != -1);
assert(close(source) != -1);
assert(close(dest) != -1);
return EXIT_SUCCESS;
}
which gives basically mostly system time as expected:
real 0m2.175s
user 0m0.001s
sys 0m1.476s
I was also curious to see if time
would distinguish between syscalls of different processes, so I tried:
time ./sendfile.out sendfile.in1.tmp sendfile.out1.tmp &
time ./sendfile.out sendfile.in2.tmp sendfile.out2.tmp &
And the result was:
real 0m3.651s
user 0m0.000s
sys 0m1.516s
real 0m4.948s
user 0m0.000s
sys 0m1.562s
The sys time is about the same for both as for a single process, but the wall time is larger because the processes are competing for disk read access likely.
So it seems that it does in fact account for which process started a given kernel work.
Bash source code
When you do just time <cmd>
on Ubuntu, it use the Bash keyword as can be seen from:
type time
which outputs:
time is a shell keyword
So we grep source in the Bash 4.19 source code for the output string:
git grep '"user\b'
which leads us to execute_cmd.c function time_command
, which uses:
gettimeofday()
and getrusage()
if both are availabletimes()
otherwiseall of which are Linux system calls and POSIX functions.
GNU Coreutils source code
If we call it as:
/usr/bin/time
then it uses the GNU Coreutils implementation.
This one is a bit more complex, but the relevant source seems to be at resuse.c and it does:
wait3
call if that is availabletimes
and gettimeofday
otherwiseI want to mention some other scenario when the real-time is much much bigger than user + sys. I've created a simple server which respondes after a long time
real 4.784
user 0.01s
sys 0.01s
the issue is that in this scenario the process waits for the response which is not on the user site nor in the system.
Something similar happens when you run the find
command. In that case, the time is spent mostly on requesting and getting a response from SSD.
In very simple terms, I like to think about it like this:
real
is the actual amount of time it took to run the command (as if you had timed it with a stopwatch)
user
and sys
are how much 'work' the CPU
had to do to execute the command. This 'work' is expressed in units of time.
Generally speaking:
user
is how much work the CPU
did to run to run the command's codesys
is how much work the CPU
had to do to handle 'system overhead' type tasks (such as allocating memory, file I/O, ect.) in order to support the running commandSince these last two times are counting 'work' done, they don't include time a thread might have spent waiting (such as waiting on another process or for disk I/O to finish).
real
, however, is a measure of actual runtime and not 'work', so it does include any time spent waiting.
time
, have it do something that will take at least a second. – Peter Cordestime
is a bash keyword. So typingman time
is not giving you a man page for the bashtime
, rather it is giving the man page for/usr/bin/time
. This has tripped me up. – irritable_phd_syndrome